Question

Let the function \(f:[1,\infin)→\R\) be defined by
\(f(t) = \begin{cases}     (-1)^{n+1}2, & \text{if } t=2n-1,n\in\N, \\     \frac{(2n+1-t)}{2}f(2n-1)+\frac{(t-(2n-1))}{2}f(2n+1) & \text{if } 2n-1<t<2n+1,n\in\N. \end{cases}\)
Define \(g(x)=\int\limits_{1}^{x}f(t)dt,x\in(1,\infin).\) Let α denote the number of solutions of the equation g(x) = 0 in the interval (1, 8] and \(β=\lim\limits_{x→1+}\frac{g(x)}{x-1}\). Then the value of α + β is equal to _____.

Answer 1

Correct Answer: 5

Piecewise Definition of f(t) 

The piecewise definition of \( f(t) \) is given as:

\[ f(t) = \begin{cases} 2, & t = 1 \\ 4 - 2t, & 1 < t < 3 \\ -2, & t = 3 \\ -8 - 2t, & 3 < t < 5 \\ \text{and so on.} \end{cases} \]

Step 1: Evaluate \( g(x) \)

The integral \( g(x) \) is defined as:

\[ g(x) = \int_1^x f(t) \, dt \]

From the structure of \( f(t) \), the integral satisfies:

\[ g(x) = 0 \quad \text{when} \quad x = 3, 5, 7, \dots \]

Thus, we identify \( \alpha = 3 \), the first point where \( g(x) = 0 \) after \( x = 1 \).

Step 2: Determine \( \beta \)

We calculate the limit:

\[ \beta = \lim_{x \to 1^+} \frac{g(x)}{x - 1} \]

From the Fundamental Theorem of Calculus:

\[ \beta = f(1) = 2 \]

Thus, \( \beta = 2 \).

Step 3: Final Result

The sum is:

\[ \alpha + \beta = 3 + 2 = 5 \]

Final Answer:

• \( \alpha + \beta = 5 \)